Mumford's Biextensions
Seminar 3:00pm ness: ok, so let's begin 3:01pm ness: So recall that a holomorphic function can be (locally) expanded as a power series: 3:02pm ness: f(z) = sum_{n=0}^{infty} c_n (z-a)^n, where c_n = f^(n)(a)/n! 3:02pm ness: This is just taylor's theorem in the complex case 3:03pm ness: But a function described this way obviously cannot have singularities in the region (disk) of convergence 3:03pm ness: so taylor series are of limited use if we want to study functions that contain poles but are "nice" otherwise 3:04pm ness: consider therefore the following straightforward extension: 3:04pm ness: f(z) = sum_{n=-infty}^{infty} c_n (z-a)^n, for some c_n, where it converges 3:04pm ness: This is called a laurent series expansion. 3:05pm ness: We will talk a bit about properties of laurent series 3:05pm ness: Throughout, let R^* denote R_0^+ u {infty}, i.e. the non-negative reals or infinity 3:06pm ness: this is purely symbolic, with the obvious ordering, but no more structure 3:06pm ness: We might first of all wonder: what is the region of convergence of a particular laurent series? 3:07pm ness: There is a relatively simple answer: there exist r_1 < r_2 in R^* such that the laurent series converges inside the annullus D(a, r_1, r_2) = {z in C | r_1 < |z-a| < r_2}, and diverges outside of it. 3:08pm ness: As in the taylor series case, we cannot know in advance for z on the border 3:09pm ness: I won't prove this here, but it can be proved using much the same ideas that are used in the analogous theorem about taylor series, and there are similar lim sup formulas to determine r_1 and r_2 3:09pm ness: There are a few special cases: 3:10pm ness: if c_n = 0 for all n < some n_0 (where n_0 < 0), then obviously r_1 = 0 3:11pm ness: further, if we have c_n = 0 for all n<=0 we are left with a holomorphic function and thus we know that not only r_1 = 0 but also that the series converges at a at well 3:12pm ness: Now for a bit of terminology: the negative powers part of the laurent series (i.e. the c_{-1}/z + c_{-2}/z^2 + ...) is called the prinzipal part 3:12pm ness: er principal part 3:12pm ness: so we can restate the above as: if the laurent series has a finite principal part, then r_1 = 0 3:13pm ness: We might ask: what else can we do with laurent series? 3:13pm ness: There is one nice application: we can classify singularities 3:13pm ness: We can say (indeed, define) the following: 3:13pm ness: If the principal part vanishes, then we have a removable singularity 3:14pm ness: If c_n = 0 for all n < -n_0 < 0, we say that the function has an n-th order pole 3:15pm ness: and finally, if none of the above holds, the singularity is said to be essential 3:16pm ness: Now of course there is a (superficially) more general way to classify singularities: Say we have a function f(z), not necessarily given as a laurent series expansion 3:16pm ness: We can consider the limit lim_{z->a) (z-a)^n f(z), for n >= 0 3:17pm ness: If the limit exists for all n (>=), then we say f has a removable singularity 3:17pm ness: if the limit exists for all n >= n_0, we say that the function has an nth order pole 3:18pm ness: and if the limit diverges for all n, we say that the function has an essential singularity at a 3:19pm ness: now that I use the same terminology here already suggests this: if a function has a laurent series expansion at a with r_1 = 0, then the two ways of classifying the singularities are *equivalent* 3:19pm ness: We will need this later 3:20pm ness: There is a formula to compute c_n (in fact it looks exactly like the cauchy formula expression for the c_n of a taylor series), but I won't talk about it here because we don't need it 3:21pm ness: So much for a detour in more or less abstract series expansions 3:21pm ness: You may well ask how all of this is related in any direct way to the topics covered last time 3:22pm ness: but there is a relation, which you will see shortly 3:22pm ness: Recall that a meromorphic function f(z) is defined as the quotient of two holomorphic functions: f(z) = g(z)/h(z) 3:23pm ness: We will see that all meromorphic functions have laurent series expansions with r_1 = 0 around all a (in their region of holomorphy) 3:24pm ness: To show this, we need a "classification of roots" 3:24pm ness: Let f(z) be a holomorphic function, and consider L = lim_{z->a} f(z)/(z-a)^(n+1), for n > 0 3:25pm ness: If L diverges for all n, we might call this a "removable root", or "no root at all" (quotation marks because this is no standard terminology) 3:25pm ness: If L=0 for na} f(z)/(z-a)^n = 0 for all n (by the definition of "essential root") 3:29pm ness: so we see that all the coefficients of the power series expansion must be zero, thus f vanishes identically 3:29pm ness: a contradiction! 3:30pm ness: Equipped with that, we can show the relation between meromorphic functions and laurent series: 3:30pm ness: lot f(z)=g(z)/h(z) where g and h are holomorphic (i.e. f is meromorphic). 3:31pm ness: Let a be in C. Then (by the above) h has at most a singularity of finite multiplicity at a, say of multiplicity n 3:31pm ness: But then phi(z) = (z-a)^n g(z)/h(z) = (z-a)^n f(z) must be holomorphic 3:32pm ness: So it has a (local) power series expansion around a 3:32pm ness: and it follows that f has a local laurent series expansion (around a) with finite principle part! 3:33pm ness: But the taylor (laurent) cofficients are arbitrary (as are g and h), so we see the following: 3:33pm ness: Meromorphic functions (ratios of holomorphic functions) are exactly the functions that have local laurent expansions with finite principal part. 3:34pm ness: Now we can go back to the residue theorem 3:35pm ness: Recall that we had shown that oint_{Gamma) f(z) dz = 2 pi i sum_a ind_{Gamma}(a) res_a f 3:36pm ness: where f is meromorphic the sum is over all the enclosed singularities (poles), Ind_{Gamma}(a) is the "winding number" of Gamma around a (recall the picture! figure 6) (read ahead for correction)3:36pm ness: and where we defined Res_a f = 2 pi i oint_C f(z) dz, where C is a closed circle around a enclosing no other singularities, oriented in the counterclockwise sense 3:37pm ness: Now we can plug in our new knowledge on laurent series here: We know that, locally, f(z) = sum_{n=-n_0}^{infty} c_n (z-a)^n 3:38pm ness: We might ask: does this detailed knowledge (assume for the sake of argument we know the correct c_n) help us to evaluate the residue? 3:38pm ness: Indeed, it does: 3:39pm ness: First of all, we see that f(z) = sum_{n=-n_0}^{-1} c_n (z-a)^n + sum_{n=0}^{infty} c_n (z-a)^n, but the second sum is actually a holomorphic function! 3:40pm ness: So wee see that all the c_n with n >= 0 cannot possibly affect the residue 3:40pm ness: Now let's look at the c_{-1} coefficient: 3:41pm ness: oh 3:41pm ness: I just saw I mistyped the definition of residue above 3:41pm ness: Res_a f = 1/(2 pi i) oint_C f(z) dz is the correct version 3:42pm ness: so the c_{-1} term contributes 1/(2 pi i) oint_C c_{-1}/z dz to the residue 3:43pm ness: but we have evaluated this integral already, in fact it was the first contour integral we ever evaluated! We know that oint_C dz/z = 2 pi i 3:43pm ness: so the c_{-1} term contributes just c_{-1} to the residue (now you see why the definition has the superficially useless 1/(2 pi i) factor!) 3:44pm ness: Now let's look at the c_n left (n < -1) 3:45pm ness: To find the contributions to the residue, we evaluate c_{-n} oint_C dz/z^n 3:46pm ness: We can do this by parametrizing the circle as e^(it), this becomes int_0^{2 pi} 1/e^(itn) * i e^it dt which is easily evaluated to be zero! 3:46pm ness: So the c_{-1} term is the *only* one contributing to the residue 3:47pm ness: in fact, we could have defined Res_a f = c_{-1}, where f(z) = sum_{n=-n_0}^{infty} c_n (z-a)^n in the first place 3:48pm ness: So we see that evaluating integrals using the residue theorem boils down to finding the c_{-1} coefficient 3:48pm ness: "tricks" for doing so are often summarized as calculus of residues 3:49pm ness: Let's start with the obvious: Res_a (f+g) = Res_a f + Res_a g, and Res_a (c*f) = c*Res_a f where c is a constant 3:50pm ness: Now of course we can always find Res_a f by expanding f into a laurent series (perhaps only partially, so that we see the c_{-1} coefficient), probably using clever expansions of the functions f is built up from 3:51pm ness: But there is one more, particularly useful formula for evaluating Res_a f, in the case that we know f to have an nth order pole at a: 3:51pm ness: assume thet f has an nth order pole at a, i.e. locally f(z) = sum_{k=-n} c_k (z-a)^k 3:52pm ness: But then (z-a)^n f(z) is holomorphic, so (again locally) (z-a)^n f(z) = sum_{k=0}^{infty} b_k (z-a)^k, where b_k = 1/k! lim_{z->a} d^k/dz^k f(z) 3:53pm ness: (The formula for b_k might look daunting, but it is just the usual c_n = f^(n)/n! for this case) 3:53pm ness: (er f^(n)(a)/n!) 3:54pm ness: But this gives us another laurent expansion of f(z), by dividing the obtained expansion of (z-a)^n f(z) by (z-a)^n 3:54pm ness: So f(z) = sum_{k=-n}^{infty} b_k (z-a)^{k-n} 3:55pm ness: and, as the regions of convergence must overlap, we can match coefficients to yield c_{-1} = b_{n-1} 3:55pm ness: So we get 3:55pm ness: Res_a f = 1/(n-1)! lim_{z->a} d^{n-1}/dz^{n-1} f(z) 3:56pm ness: Again, this might look daunting at first, but it is a very convenient way to evaluate certain integrals 3:56pm ness: To demonstrate this (and other things I talked of in the other two seminars), let me show you how the machinery developed can be used to evaluate complicated real integrals 3:57pm ness: Let's try to evaluate I = int_{-infty}^{infty} dx/(x^2+x+1) 3:57pm ness: This looks quite hard if we try it using traditional techniques (seems to be an arctangent in somewhere, but nothing obvious, at least to me) 3:57pm ness: We extend this to a contour integral: 3:58pm ness: Consider I' = oint_C dz/(z^2+z +1), where the contour is a semicircle as indicated in figure 7 3:59pm ness: The semicircle is oriented counterclockwise and the radius should tend to infinity 3:59pm ness: I claim that I' = I 4:00pm ness: Indeed, this is because lim_{|z| => infty} z*f(z) = 0 4:00pm ness: We can clearly see that our integrand satisfies this relation, but how does this help? 4:01pm ness: Now, it can be shown that |int_C f(z) dz| <= int_C |f(z)| |dz| <= L * M, where L is the length of the contour and M is a bound of |f(z)| on the contour. This is known as the standard estimate 4:03pm ness: Applying this to a (semi-circle) we find that |int_C f(z) dz| <= pi * R * M. But R*M goes to zero as R (=|z|) goes to infinity (by assumption) 4:03pm ness: so we see: I = I' 4:03pm ness: We have thus "reduced" a real integral to an integral around a closed contour in the complex plane, and we can use the residue theorem 4:04pm ness: To do so, we must first find the poles of the integrand 4:04pm ness: we solve z^2+z+1 = 0, find z_{1,2} = (-1 +- isqrt(3))/2 4:04pm ness: but only z_1 is inside the contour of integration (effectively the upper half plane) 4:05pm ness: so I = 2 pi i Res_{z_1} 1/(z^2+z+1) 4:05pm ness: But 1/(z^2+z+1) = 1/(z-z_1)*1/(z-z_2), so the pole is simple, and by the above formula for residues, 4:06pm ness: Res_{z_1} 1/(z^2+z+1) = lim_{z->z_1} (z-z_1)/(z^2+z+1) = 1/(z-z_2) evaluated at z=z_1, = 1/(i sqrt(3)) 4:07pm ness: Thus we find I' = I = 2pi/sqrt(3), almost without thinking! 4:07pm ness: Ok. 4:07pm ness: So much for the residue calculus 4:07pm ness: I will stop here 4:08pm ness: There is still plenty of stuff to explore in the wonderful world of complex analysis 4:08pm ness: Are there any questions? 4:09pm ness: Hehe, all bored away Category:Seminar Category:Mumford's Biextensions